Directory UMM :Journals:Journal_of_mathematics:EJQTDE:
Electronic Journal of Qualitative Theory of Differential Equations
2011, No. 3, 1-15; http://www.math.u-szeged.hu/ejqtde/
Periodic boundary value problems for nonlinear impulsive
fractional differential equation
Xiaojing Wang ∗ and Chuanzhi Bai
Department of Mathematics, Huaiyin Normal University, Huaian, Jiangsu 223300, P R China
Abstract
In this paper, we investigate the existence and uniqueness of solution of the periodic boundary value
problem for nonlinear impulsive fractional differential equation involving Riemann-Liouville fractional
derivative by using Banach contraction principle.
Keywords: Riemann-Liouville fractional derivative; Periodic boundary value problem; Impulsive
1. Introduction
This paper deals with the existence of solutions for nonlinear impulsive fractional differential
equation with periodic boundary conditions
D α u(t) − λu(t) = f (t, u(t)),
t ∈ (0, 1] \ {t1 },
0 < α ≤ 1,
(1.1)
lim t1−α u(t) = u(1),
(1.2)
lim (t − t1 )1−α (u(t) − u(t1 )) = I(u(t1 )),
(1.3)
t→0+
t→t+
1
where D α is the standard Riemann-Liouville fractional derivative, λ ∈ R, 0 < t1 < 1, I ∈
C(R, R), f is continuous at every point (t, u) ∈ [0, 1] × R.
For clarity and brevity, we restrict our attention to BVPs with one impulse, the difference
between the theory of one or an arbitrary number of impulses is quite minimal.
In [1], the author investigated the existence and uniqueness of solution to initial value
problems for a class of fractional differential equations
D α u(t) = f (t, u(t)),
t ∈ (0, T ]
t1−α u(t)|t=0 = u0 ,
0 < α < 1,
(1.4)
(1.5)
by using the method of upper and lower solutions and its associated monotone iterative.
∗
E-mail address: wangxj2010106@sohu.com
EJQTDE, 2011 No. 3, p. 1
In [2], the existence and uniqueness of solution of the following fractional differential equation with periodic boundary value condition
D δ u(t) − λu(t) = f (t, u(t)),
t ∈ (0, 1]
0 < δ < 1,
(1.6)
lim t1−δ u(t) = u(1),
(1.7)
t→0+
was discussed by using the fixed point theorem of Schaeffer and the Banach contraction principle.
Differential equation with fractional order have recently proved valuable tools in the modeling of many phenomena in various fields of science and engineering [3-7]. There has also been
a significant theoretical development in fractional differential equations in recent years, see for
examples [8-19]. Recently, many researchers paid attention to existence result of solution of
the initial value problem and boundary value problem for fractional differential equations, such
as [20-25].
Impulsive differential equations are now recognized as an excellent source of models to
simulate process and phenomena observed in control theory, physics, chemistry, population
dynamics, biotechnology, industrial robotic, optimal control, etc. [26,27]. Periodic boundary
value for impulsive differential equation has drawn much attention, see [28-31]. Anti-periodic
problems constitute an important class of boundary value problems and have recently received
considerable attention. The recent results on anti-periodic BVPs or impulsive anti-periodic
BVPs of fractional differential equations can be found in [32-35]. But till now, the theory
of boundary value problems for nonlinear fractional differential equations is still in the initial
stages. For some recent work on impulsive fractional differential equations, see [36-41] and the
references therein.
To the best of the authors knowledge, no one has studied the existence of solutions for BVP
(1.1)-(1.3). The purpose of this paper is to study the existence and uniqueness of solution of
the periodic boundary value problem for nonlinear impulsive fractional differential equation
involving Riemann-Liouville fractional derivative by using Banach contraction principle.
2. Preliminaries
In this section, we introduce notations, definitions, and preliminary facts which are used
throughout this paper. Let C(a, b] (C[a, b]) be the Banach space of all continuous real functions
defined on (a, b] ([a, b]).
EJQTDE, 2011 No. 3, p. 2
In order to define the solution of (1.1)-(1.3) we shall consider the space
P Cr [0, 1] = {x : tr x|[0,t1 ] ∈ C[0, t1 ], (t − t1 )r x|(t1 ,1] ∈ C(t1 , 1],
−
there exist lim (t − t1 )r x(t) and x(t−
1 ) with x(t1 ) = x(t1 )},
t→t+
1
where constant 0 ≤ r < 1. It is easy to check that the space P Cr [0, 1] is a Banach space with
norm
kxkr = max{sup{tr |x(t)| : t ∈ [0, t1 ]}, sup{(t − t1 )r |x(t)| : t ∈ (t1 , 1]}}.
Remark 2.1. If r = 0, then the definition of P Cr [0, 1] reduces to the following
P C[0, 1] = {x : x|[0,t1 ] ∈ C[0, t1 ], x|(t1 ,1] ∈ C(t1 , 1],
+
−
there exist x(t−
1 ) and x(t1 ) with x(t1 ) = x(t1 )}.
And the space P C[0, 1] is a Banach space with norm
kxkP C = sup{|x(t)| : t ∈ [0, 1]}.
Lemma 2.1. The linear impulsive boundary value problem
D α u(t) − λu(t) = σ(t),
t ∈ (0, 1] \ {t1 },
(2.1)
lim t1−α u(t) = u(1),
(2.2)
lim (t − t1 )1−α (u(t) − u(t1 )) = a1 ,
(2.3)
t→0+
t→t+
1
where λ, a1 ∈ R are constants and σ ∈ C[0, 1], has a unique solution u ∈ P C1−α [0, 1] given by
u(t) =
1
Z
Gλ,α (t, s)σ(s)ds + Γ(α)Gλ,α (t, t1 )a1 ,
0
(2.4)
where
Gλ,α (t, s)
=
Γ(α)Eα,α (λtα )Eα,α (λ(1−s)α )tα−1 (1−s)α−1
+
1−Γ(α)Eα,α (λ)
Γ(α)Eα,α (λtα )Eα,α (λ(1−s)α )tα−1 (1−s)α−1
,
1−Γ(α)Eα,α (λ)
λk
k=0 Γ((k+1)α)
P∞
and Eα,α (λ) =
(t − s)α−1 Eα,α (λ(t − s)α ), 0 ≤ s ≤ t ≤ 1,
0 ≤ t < s ≤ 1,
is Mittag-Leffler function (see [6]).
Proof. By Theorem 3.2 in [2], we have that
u
¯(t) =
Z
0
1
Gλ,α (t, s)σ(s)ds,
t ∈ (0, 1] \ {t1 }
(2.5)
EJQTDE, 2011 No. 3, p. 3
is a unique solution of the linear problem (2.1)-(2.2), and u¯ ∈ P C1−α [0, 1]. Set
w(t) = Gλ,α (t, t1 )Γ(α)a1 ,
t ∈ [0, 1].
For each t < t1 , we have
D α w(t) =
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α α−1
D (t
Eα,α (λtα ))
1 − Γ(α)Eα,α (λ)
∞
λi−1 αi−1
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α X
D
t
=
.
1 − Γ(α)Eα,α (λ)
Γ(αi)
i=1
!
Using the identities
D α tµ =
Γ(µ + 1) µ−α
t
(µ > −1),
Γ(µ + 1 − α)
Dα tα−1 = 0,
we get
D α w(t) =
∞
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 X
λi−1
tα(i−1)−1
1 − Γ(α)Eα,α (λ)
Γ(α(i
−
1))
i=2
=λ
∞
(−λ)i−1 αi−1
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 X
t
1 − Γ(α)Eα,α (λ)
Γ(αi)
i=1
=λ
∞
(λtα )i
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α−1 X
t
1 − Γ(α)Eα,α (λ)
Γ(αi + α)
i=0
=λ
Γ(α)Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 tα−1 Eα,α (λtα )
Γ(α)a1
1 − Γ(α)Eα,α (λ)
= λGα,α (t, t1 )Γ(α)a1 = λw(t),
t < t1 .
Similarly, we can obtain that
D α w(t) − λw(t) = 0,
t > t1 .
Thus, we have Dα w(t) − λw(t) = 0 for t ∈ (0, 1] \ {t1 }. Moreover, we have
lim t1−α w(t) − w(1)
t→0+
= lim+ t1−α
t→0
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α−1
t
Eα,α (λtα )
1 − Γ(α)Eα,α (λ)
!
Γ(α)Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 Eα,α (λ)
−
+ (1 − t1 )α−1 Eα,α (λ(1 − t1 )α ) Γ(α)a1
1 − Γ(α)Eα,α (λ)
=
Γ(α)Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 (1 − t1 )α−1 Eα,α (λ(1 − t1 )α )Γ(α)a1
−
1 − Γ(α)Eα,α (λ)
1 − Γ(α)Eα,α (λ)
EJQTDE, 2011 No. 3, p. 4
= 0,
and
lim+ (t − t1 )1−α (w(t) − w(t1 ))
t→t1
= lim (t − t1 )1−α · (t − t1 )α−1 Eα,α (λ(t − t1 )α )Γ(α)a1
t→t+
1
α
= lim Eα,α (λ(t − t1 ) )Γ(α)a1 = lim
t→t+
1 i=0
t→t+
1
=
1
Γ(α) Γ(α)a1
∞
X
(λ(t − t1 )α )i
Γ(αi + α)
Γ(α)a1
= a1 .
In consequence,
u(t) = u
¯(t) + w(t) =
Z
1
0
Gλ,α (t, s)σ(s)ds + Γ(α)Gλ,α (t, t1 )a1
is the solution of problem (2.1)-(2.3).
Next we prove that the solution of BVP (2.1)-(2.3) is unique. Suppose that u1 , u2 ∈
P C1−α [0, 1] are two solutions of BVP (2.1)-(2.3). Let v = u1 − u2 , then we have
D α v(t) − λv(t) = 0,
t ∈ (0, 1] \ {t1 },
(2.6)
lim t1−α v(t) − v(1) = 0,
(2.7)
lim (t − t1 )1−α (v(t) − v(t1 )) = 0.
(2.8)
t→0+
t→t+
1
By (2.5) and (2.6)-(2.7), we get that v(t) ≡ 0 for any t ∈ (0, 1] \ {t1 }. Since v ∈ P C1−α [0, 1],
v(t1 ). On the other hand, limt→t− t1−α v(t) = 0. Thus, we
we have limt→t− t1−α v(t) = t1−α
1
1
1
obtain that v(t1 ) = 0. Hence, u1 (t) = u2 (t) for each t ∈ (0, 1]. Moreover, by (2.7), we
have limt→0+ t1−α v(t) = v(1) = 0, which implies that limt→0+ t1−α u1 (t) = limt→0+ t1−α u2 (t).
Therefore, u1 = u2 .
Remark 2.2. For α = 1, Lemma 2.1 reduces to the one for a first order linear impulsive
boundary value problem
3. Main results
Lemma 3.1. Suppose that
(H1 ) there exist positive constants M and m such that
EJQTDE, 2011 No. 3, p. 5
|f (t, u)| ≤ M,
|I(u)| ≤ m,
∀t ∈ [0, 1], u ∈ R,
(3.1)
holds, then the operator A : P C1−α [0, 1] → P C1−α [0, 1] is well defined.
Proof. Define the operator A as follows :
(Ax)(t) :=
Z
1
0
Gλ,α (t, s)f (s, x(s))ds + Γ(α)Gλ,α (t, t1 )I(x(t1 )).
(3.2)
By Lemma 2.1, it is easy to see that a function x is a solution to (1.1)-(1.3) if and only
if x is a fixed point of A. Using (H1 ), we check that t1−α (Au)(t)|[0,t1 ] ∈ C[0, t1 ] and (t −
t1 )1−α (Au)(t)|(t1 ,1] ∈ C(t1 , 1]. Here, we only prove that (t − t1 )1−α (Au)(t)|(t1 ,1] ∈ C(t1 , 1].
Similarly, we can prove that t1−α (Au)(t)|[0,t1 ] ∈ C[0, t1 ]. For every u ∈ P C1−α [0, 1], for any
τ1 , τ2 ∈ (t1 , 1] with τ1 < τ2 , we have
(τ1 − t1 )1−α (Au)(τ1 ) − (τ2 − t1 )1−α (Au)(τ2 )
Z
≤ (τ1 − t1 )1−α
1
0
1−α
Gλ,α (τ1 , s)f (s, u(s))ds − (τ2 − t1 )
Z
0
1
Gλ,α (τ2 , s)f (s, u(s))ds
+ (τ1 − t1 )1−α Γ(α)Gλ,α (τ1 , t1 )I(u(t1 )) − (τ2 − t1 )1−α Γ(α)Gλ,α (τ2 , t1 )I(u(t1 ))
Z
1 Γ(α)E (λ(1 − s)α )(1 − s)α−1
α,α
=
0
1 − Γ(α)Eα,α (λ)
×[(τ1 − t1 )1−α Eα,α (λτ1α )τ1α−1 − (τ2 − t1 )1−α Eα,α (λτ2α )τ2α−1 ]f (s, u(s))ds
+
Z
τ1
0
[(τ1 − t1 )1−α (τ1 − s)α−1 Eα,α (λ(τ1 − s)α )
−(τ2 − t1 )1−α (τ2 − s)α−1 Eα,α (λ(τ2 − s)α )]f (s, u(s))ds
−
Z
τ2
τ1
≤
(τ2 − t1 )1−α (τ2 − s)α−1 Eα,α (λ(τ2 − s)α f (s, u(s))ds
+ [(τ1 − t1 )1−α Gλ,α (τ1 , t1 ) − (τ2 − t1 )1−α Gλ,α (τ2 , t1 )]Γ(α)I(u(t1 ))
Γ(α)M
|(τ1 − t1 )1−α Eα,α (λτ1α )τ1α−1 − (τ2 − t1 )1−α Eα,α (λτ2α )τ2α−1 |
|1 − Γ(α)Eα,α (λ)|
×
Z
0
1
Eα,α (λ(1 − s)α )(1 − s)α−1 ds
+M ((τ2 − t1 )1−α − (τ1 − t1 )1−α )
+M (τ2 − t1 )1−α
Z
0
τ1
Z
τ1
0
(τ1 − s)α−1 Eα,α (λ(τ1 − s)α )ds
|(τ1 − s)α−1 Eα,α (λ(τ1 − s)α ) − (τ2 − s)α−1 Eα,α (λ(τ2 − s)α )|ds
EJQTDE, 2011 No. 3, p. 6
+M (τ2 − t1 )1−α
+
τ2
Z
τ1
(τ2 − s)α−1 Eα,α (λ(τ2 − s)α )ds
Γ2 (α)mEα,α (λ(1 − t1 )α )(1 − t1 )1−α
|1 − Γ(α)Eα,α (λ)|
×|(τ1 − t1 )1−α τ1α−1 Eα,α (λτ1α ) − (τ2 − t1 )1−α τ2α−1 Eα,α (λτ2α )|
+|Eα,α (λ(τ1 − t1 )α ) − Eα,α (λ(τ2 − t1 )α )|
→ 0,
as τ1 → τ2 ,
since
Z
1
Z
τ1
Z
τ2
Z
τ1
0
Eα,α (λ(1 − s)α )(1 − s)α−1 ds = Eα,α+1 (λ) ≤ Eα,α+1 (|λ|),
0
τ1
and
0
≤
Eα,α (λ(τ1 − s)α )(τ1 − s)α−1 ds = τ1α Eα,α+1 (λτ1α ) ≤ τ1α Eα,α+1 (|λ|τ1α ),
Eα,α (λ(τ2 − s)α )(τ2 − s)α−1 ds = (τ2 − τ1 )α Eα,α+1 (λ(τ2 − τ1 )α ),
|(τ1 − s)α−1 Eα,α (λ(τ1 − s)α ) − (τ2 − s)α−1 Eα,α (λ(τ2 − s)α )|ds
∞
X
i=0
|λ|i
Γ(αi + α)
Z
τ1
0
(τ1 − s)αi+α−1 − (τ2 − s)αi+α−1 ds
tend to zero as τ1 → τ2 by (4.13) and (4.14) in [2].
So, (t − t1 )1−α (Au)(t)|(t1 ,1] ∈ C(t1 , 1]. On the other hand, we have limt→t− (Au)(t) and
1
lim (t − t1 )1−α (Au)(t)
t→t+
1
= lim (t − t1 )1−α
t→t+
1
+
Z
0
t
"Z
1
0
Γ(α)Eα,α (λtα )Eα,α (λ(1 − s)α )tα−1 (1 − s)α−1
f (s, u(s))ds
1 − Γ(α)Eα,α (λ)
(t − s)α−1 Eα,α (λ(t − s)α )f (s, u(s))ds
+Γ(α)
Γ(α)Eα,α (λtα )Eα,α (λ(1 − t1 )α )tα (1 − t1 )α−1
1 − Γ(α)Eα,α (λ)
i
+(t − t1 )α−1 Eα,α (λ(t − t1 )α ) I(u(t1 ))
= lim (t − t1 )1−α · Γ(α)(t − t1 )α−1 Eα,α (λ(t − t1 )α )I(u(t1 ))
t→t+
1
EJQTDE, 2011 No. 3, p. 7
= I(u(t1 ))
exist. Thus, A : P C1−α [0, 1] → P C1−α [0, 1].
For convenience, set
M1 :=
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1 tα1
(Γ(α))2
+ tα1 Eα,α (|λ|tα1 )
,
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
(3.3)
M2 :=
(Γ(α))3 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )(1 − t1 )2α−1
,
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.4)
M3 :=
M4 :=
Γ(α)(Eα,α (|λ|))2 t12α−1
|1−Γ(α)(Eα,α (λ)|α
Γ(α)(Eα,α (|λ|))2 t12α−1
|1−Γ(α)Eα,α (λ)|α
2
1
+ t2α−1
(1 − t1 )1−α Eα,α (|λ|) (Γ(α))
1
Γ(2α) , if 0 < α ≤ 2 ,
2
+ (1 − t1 )1−α Eα,α (|λ|) (Γ(α))
Γ(2α) ,
if
1
2
< α ≤ 1,
(Γ(α))3 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )tα−1
(1 − t1 )α
1
,
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.5)
(3.6)
N1 :=
(1 − t1 )α−1
(Γ(α))2 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )tα−1
1
,
|1 − Γ(α)Eα,α (λ)|
(3.7)
N2 :=
(Γ(α))2 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )t2α−2
1
+ Γ(α)Eα,α (|λ|(1 − t1 )α )tα−1
.
1
|1 − Γ(α)Eα,α (λ)|
(3.8)
Theorem 3.2. Suppose that (H1 ) and the following condition hold:
(H2 ) there exist positive constant k, and l such that
|f (t, u) − f (t, v)| ≤ k|u − v|,
|I(u) − I(v)| ≤ l|u − v|,
∀t ∈ [0, 1], u, v ∈ R.
(3.9)
Then the problem (1.1)-(1.3) has a unique solution in P C1−α [0, 1], provided that
max{k(M1 + M2 ) + lN1 , k(M3 + M4 ) + lN2 } < 1.
(3.10)
Proof. By (H1 ) and Lemma 3.1, we have A : P C1−α [0, 1] → P C1−α [0, 1]. For any x, y ∈
P C1−α [0, 1], and each t ∈ [0, t1 ], we obtain by (H2 ) that
t
1−α
|(Ax)(t) − (Ay)(t)| ≤
Z
1
0
t1−α |Gλ,α (t, s)||f (s, x(s)) − f (s, y(s))|ds
+t1−α Γ(α)|Gλ,α (t, t1 )||I(x(t1 )) − I(y(t1 ))|
≤
Z
1
0
=k
Z
t1−α |Gλ,α (t, s)|k|x(s) − y(s)|ds + t1−α Γ(α)|Gλ,α (t, t1 )|l|x(t1 ) − y(t1 )|
0
t1
t1−α |Gλ,α (t, s)|sα−1 s1−α |x(s) − y(s)|ds
EJQTDE, 2011 No. 3, p. 8
1
Z
+k
t1−α |Gλ,α (t, s)|(s − t1 )α−1 (s − t1 )1−α |x(s) − y(s)|ds
t1
+lΓ(α)t1−α |Gλ,α (t, t1 )|tα−1
t11−α |x(t1 ) − y(t1 )|
1
Z
≤ kx − yk1−α k
+k
1
Z
t
t1
1−α
t1
0
t1−α |Gλ,α (t, s)|sα−1 ds
α−1
|Gλ,α (t, s)|(s − t1 )
ds + lΓ(α)t
1−α
|Gλ,α (t, t1 )|tα−1
1
(3.11)
.
From the expression of Gλ,α (t, s), and (4.21) in [2], we have
Z
t1
t
1−α
0
|Gλ,α (t, s)|s
+
Z
t
0
≤
α−1
ds ≤
Z
t1
t
0
1−α Γ(α)Eα,α (λt
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1 α−1
ds
s
1 − Γ(α)Eα,α (λ)
t1−α (t − s)α−1 |Eα,α (λ(t − s)α )|sα−1 ds
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1
|1 − Γ(α)Eα,α (λ)|
+t1−α Eα,α (|λ|tα1 )
Z
t
Z
t1
sα−1 ds
0
(t − s)α−1 sα−1 ds
0
=
(Γ(α))2
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1 tα1
+ tα Eα,α (|λ|tα1 )
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
≤
(Γ(α))2
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1 tα1
+ tα1 Eα,α (|λ|tα1 )
= M1 .
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
(3.12)
And
Z
1
t1
t1−α |Gλ,α (t, s)|(s − t1 )α−1 ds
≤
Z
1
t
t1
≤
=
1−α Γ(α)Eα,α (λt
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1
(s − t1 )α−1 ds
1 − Γ(α)Eα,α (λ)
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )
|1 − Γ(α)Eα,α (λ)|
Z
1
t1
(1 − s)α−1 (s − t1 )α−1 ds
(Γ(α))3 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )(1 − t1 )2α−1
= M2 ,
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.13)
since
Z
1
t1
α−1
(1 − s)
α−1
(s − t1 )
ds =
Z
0
1−t1
sα−1 (1 − t1 − s)α−1 ds = (1 − t1 )2α−1
(Γ(α))2
.
Γ(2α)
(3.14)
EJQTDE, 2011 No. 3, p. 9
Moreover, we have
Γ(α)t1−α |Gλ,α (t, t1 )|tα−1
1
≤
(1 − t1 )α−1
(Γ(α))2 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )tα−1
1
= N1 .
|1 − Γ(α)Eα,α (λ)|
(3.15)
Substituting (3.12), (3.13) and (3.15) into (3.11), we get
sup t1−α |(Ax)(t) − (Ay)(t)| ≤ (k(M1 + M2 ) + N1 )kx − yk1−α .
(3.16)
t∈[0,t1 ]
On the other hand, for each x, y ∈ P C1−α [0, 1] and t ∈ (t1 , 1], we obtain that
(t − t1 )1−α |(Ax)(t) − (Ay)(t)|
≤
1
Z
0
(t − t1 )1−α |Gλ,α (t, s)||f (s, x(s)) − f (s, y(s))|ds
+(t − t1 )1−α Γ(α)|Gλ,α (t, t1 )||I(x(t1 )) − I(y(t1 ))|
≤
1
Z
0
(t − t1 )1−α |Gλ,α (t, s)|k|x(s) − y(s)|ds
+(t − t1 )1−α Γ(α)|Gλ,α (t, t1 )|l|x(t1 ) − y(t1 )|
=k
Z
t1
0
+k
(t − t1 )1−α |Gλ,α (t, s)|sα−1 s1−α |x(s) − y(s)|ds
1
Z
t1
(t − t1 )1−α |Gλ,α (t, s)|(s − t1 )α−1 (s − t1 )1−α |x(s) − y(s)|ds
+(t − t1 )1−α Γ(α)l|Gλ,α (t, t1 )|tα−1
t1−α
|x(t1 ) − y(t1 )|
1
1
Z
≤ kx − yk1−α k
+k
Z
1
t1
0
t1
(t − t1 )1−α |Gλ,α (t, s)|sα−1 ds
(t − t1 )1−α |Gλ,α (t, s)|(s − t1 )α−1 ds + lΓ(α)|Gλ,α (t, t1 )|(t − t1 )1−α tα−1
.
1
(3.17)
By (4.21) in [2], we have
Z
0
t1
(t − t1 )1−α |Gλ,α (t, s)|sα−1 ds
≤
Z
t1
Z
t
0
+
0
1−α Γ(α)Eα,α (λt
(t − t1 )
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1 α−1
ds
s
1 − Γ(α)Eα,α (λ)
(t − t1 )1−α (t − s)α−1 |Eα,α (λ(t − s)α )|sα−1 ds
EJQTDE, 2011 No. 3, p. 10
≤ (1 − t1 )1−α
Γ(α)(Eα,α (|λ|))2 tα−1
(1 − t1 )α−1
1
|1 − Γ(α)Eα,α (λ)|
+(1 − t1 )1−α Eα,α (|λ|)
Z
t
Z
t1
sα−1 ds
0
(t − s)α−1 sα−1 ds
0
=
(Γ(α))2
Γ(α)(Eα,α (|λ|))2 t2α−1
1
+ (1 − t1 )1−α Eα,α (|λ|)t2α−1
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
≤
Γ(α)(Eα,α (|λ|))2 t2α−1
1
|1 − Γ(α)Eα,α (λ)|α
t2α−1 (1 − t1 )1−α Eα,α (|λ|) (Γ(α))2 , if 0 < α ≤ 1 ,
1
2
Γ(2α)
+
2
1
(1 − t1 )1−α Eα,α (|λ|) (Γ(α)) ,
if 2 < α ≤ 1,
Γ(2α)
= M3 .
(3.18)
From (3.14), we get
Z
1
t1
(t − t1 )1−α |Gλ,α (t, s)|(s − t1 )α−1 ds
=
Z
1
t1
(t − t1 )
≤ (1 − t1 )1−α
=
1−α Γ(α)Eα,α (λt
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1
(s − t1 )α−1 ds
1 − Γ(α)Eα,α (λ)
Γ(α)Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )tα−1
1
|1 − Γ(α)Eα,α (λ)|
Z
1
t1
(s − t1 )α−1 (1 − s)α−1 ds
(Γ(α))3 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )tα−1
(1 − t1 )α
1
= M4 .
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.19)
Moreover, we have
Γ(α)|Gλ,α (t, t1 )|(t − t1 )1−α tα−1
1
≤
(Γ(α))2 Eα,α (|λ|tα )Eα,α (|λ|(1 − t1 )α )tα−1 (1 − t1 )α−1 (t − t1 )1−α tα−1
1
|1 − Γ(α)Eα,α (λ)|
+Γ(α)tα−1
Eα,α (|λ|(t − t1 )α )
1
≤
(Γ(α))2 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )t2α−2
1
|1 − Γ(α)Eα,α (λ)|
+Γ(α)Eα,α (|λ|(1 − t1 )α )tα−1
= N2 .
1
(3.20)
Substituting (3.18)-(3.20) into (3.17), we obtain
sup (t − t1 )1−α |(Ax)(t) − (Ay)(t)| ≤ (k(M3 + M4 ) + lN )kx − yk1−α .
(3.21)
t∈(t1 ,1]
EJQTDE, 2011 No. 3, p. 11
From (3.11), (3.16) and (3.21), we have
kAx − Ayk1−α
= max
(
sup t1−α |(Ax)(t) − (Ay)(t)|,
t∈[0,t1 ]
)
sup (t − t1 )1−α |(Ax)(t) − (Ay)(t)|
t∈(t1 ,1]
≤ max{k(M1 + M2 ) + lN1 , k(M3 + M4 ) + lN2 }kx − yk1−α ,
(3.22)
which implies that A is a contraction (by (3.10)). Therefore it has a unique fixed point.
4. An example
The following illustrative example will demonstrate the effectiveness of our main result.
Example 4.1. Consider the nonlinear impulsive fractional differential equation with periodic
boundary conditions as follows:
1
t2
D 0.9 u(t) − u(t) = 2
arctan u(t),
3
t + 11
t ∈ (0, 1] \ {0.5},
(4.1)
lim t0.1 u(t) = u(1),
(4.2)
t→0+
lim (t − 0.5)0.1 (u(t) − u(0.5)) =
t→0.5+
1
sin(u(0.5)).
13
(4.3)
Then BVP (4.1)-(4.3) reduces to BVP (1.1)-(1.3) with α = 0.9, λ = 13 , f (t, u) =
I(u) =
1
13
t2
t2 +11
arctan u,
sin u and t1 = 0.5.
Obviously, there exist positive constants M =
|f (t, u)| ≤ M,
|I(u)| ≤ m,
π
24 ,
m=
1
13 ,
k=
1
12
and l =
1
13
such that
t ∈ [0, 1], u ∈ R,
and
|f (t, u) − f (t, v)| ≤ k|u − v|,
|I(u) − I(v)| ≤ l|u − v|,
t ∈ [0, 1], u, v ∈ R.
So that the conditions (H1 ) and (H2 ) hold. Moreover, we have
Eα,α (λ) = E0.9,0.9
1
3
=
1
1
1
1
1
+
+ 2
+ 3
+ 4
+ ···
Γ(0.9) 3Γ(1.8) 3 Γ(2.7) 3 Γ(3.6) 3 Γ(4.5)
Thus, we obtain
E0.9,0.9
1
3
>
1
1
1
+
+ 2
= 1.3656,
Γ(0.9) 3Γ(1.8) 3 Γ(2.7)
(4.4)
and
EJQTDE, 2011 No. 3, p. 12
E0.9,0.9
1
3
1
1
1
1
+
+ 2
+
<
Γ(0.9) 3Γ(1.8) 3 Γ(2.7) 3
1
1
+ 4 + ···
3
3
3
= 1.3656 + 0.0185 = 1.3841.
(4.5)
Similarly, we have
Eα,α (|λ|tα1 )
α
= Eα,α (|λ|(1 − t1 ) ) = E0.9,0.9
1
3 · 20.9
< 1.149.
(4.6)
By (4.4)-(4.6) and simple calculation, we can obtain the estimation of M1 , M2 , M3 , M4 , N1 and
N2 (are as in (3.3)-(3.8) as follows :
M1 < 3.019,
M2 < 1.9351,
M4 < 2.4333,
N1 < 3.6153,
M3 < 4.2091,
N2 < 5.6709.
Hence, we get
max{k(M1 + M2 ) + lN1 , k(M3 + M4 ) + lN2 } = 0.9898 < 1,
that is (3.10) holds. So, it follows from Theorem 3.2 that the BVP (4.1)-(4.3) has a unique
solution in P C0.1 [0, 1].
Acknowledgements
The authors would like to thank the referee for their valuable suggestions and comments.
This work is supported by the National Natural Science Foundation of China (10771212).
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(Received September 5, 2010)
EJQTDE, 2011 No. 3, p. 15
2011, No. 3, 1-15; http://www.math.u-szeged.hu/ejqtde/
Periodic boundary value problems for nonlinear impulsive
fractional differential equation
Xiaojing Wang ∗ and Chuanzhi Bai
Department of Mathematics, Huaiyin Normal University, Huaian, Jiangsu 223300, P R China
Abstract
In this paper, we investigate the existence and uniqueness of solution of the periodic boundary value
problem for nonlinear impulsive fractional differential equation involving Riemann-Liouville fractional
derivative by using Banach contraction principle.
Keywords: Riemann-Liouville fractional derivative; Periodic boundary value problem; Impulsive
1. Introduction
This paper deals with the existence of solutions for nonlinear impulsive fractional differential
equation with periodic boundary conditions
D α u(t) − λu(t) = f (t, u(t)),
t ∈ (0, 1] \ {t1 },
0 < α ≤ 1,
(1.1)
lim t1−α u(t) = u(1),
(1.2)
lim (t − t1 )1−α (u(t) − u(t1 )) = I(u(t1 )),
(1.3)
t→0+
t→t+
1
where D α is the standard Riemann-Liouville fractional derivative, λ ∈ R, 0 < t1 < 1, I ∈
C(R, R), f is continuous at every point (t, u) ∈ [0, 1] × R.
For clarity and brevity, we restrict our attention to BVPs with one impulse, the difference
between the theory of one or an arbitrary number of impulses is quite minimal.
In [1], the author investigated the existence and uniqueness of solution to initial value
problems for a class of fractional differential equations
D α u(t) = f (t, u(t)),
t ∈ (0, T ]
t1−α u(t)|t=0 = u0 ,
0 < α < 1,
(1.4)
(1.5)
by using the method of upper and lower solutions and its associated monotone iterative.
∗
E-mail address: wangxj2010106@sohu.com
EJQTDE, 2011 No. 3, p. 1
In [2], the existence and uniqueness of solution of the following fractional differential equation with periodic boundary value condition
D δ u(t) − λu(t) = f (t, u(t)),
t ∈ (0, 1]
0 < δ < 1,
(1.6)
lim t1−δ u(t) = u(1),
(1.7)
t→0+
was discussed by using the fixed point theorem of Schaeffer and the Banach contraction principle.
Differential equation with fractional order have recently proved valuable tools in the modeling of many phenomena in various fields of science and engineering [3-7]. There has also been
a significant theoretical development in fractional differential equations in recent years, see for
examples [8-19]. Recently, many researchers paid attention to existence result of solution of
the initial value problem and boundary value problem for fractional differential equations, such
as [20-25].
Impulsive differential equations are now recognized as an excellent source of models to
simulate process and phenomena observed in control theory, physics, chemistry, population
dynamics, biotechnology, industrial robotic, optimal control, etc. [26,27]. Periodic boundary
value for impulsive differential equation has drawn much attention, see [28-31]. Anti-periodic
problems constitute an important class of boundary value problems and have recently received
considerable attention. The recent results on anti-periodic BVPs or impulsive anti-periodic
BVPs of fractional differential equations can be found in [32-35]. But till now, the theory
of boundary value problems for nonlinear fractional differential equations is still in the initial
stages. For some recent work on impulsive fractional differential equations, see [36-41] and the
references therein.
To the best of the authors knowledge, no one has studied the existence of solutions for BVP
(1.1)-(1.3). The purpose of this paper is to study the existence and uniqueness of solution of
the periodic boundary value problem for nonlinear impulsive fractional differential equation
involving Riemann-Liouville fractional derivative by using Banach contraction principle.
2. Preliminaries
In this section, we introduce notations, definitions, and preliminary facts which are used
throughout this paper. Let C(a, b] (C[a, b]) be the Banach space of all continuous real functions
defined on (a, b] ([a, b]).
EJQTDE, 2011 No. 3, p. 2
In order to define the solution of (1.1)-(1.3) we shall consider the space
P Cr [0, 1] = {x : tr x|[0,t1 ] ∈ C[0, t1 ], (t − t1 )r x|(t1 ,1] ∈ C(t1 , 1],
−
there exist lim (t − t1 )r x(t) and x(t−
1 ) with x(t1 ) = x(t1 )},
t→t+
1
where constant 0 ≤ r < 1. It is easy to check that the space P Cr [0, 1] is a Banach space with
norm
kxkr = max{sup{tr |x(t)| : t ∈ [0, t1 ]}, sup{(t − t1 )r |x(t)| : t ∈ (t1 , 1]}}.
Remark 2.1. If r = 0, then the definition of P Cr [0, 1] reduces to the following
P C[0, 1] = {x : x|[0,t1 ] ∈ C[0, t1 ], x|(t1 ,1] ∈ C(t1 , 1],
+
−
there exist x(t−
1 ) and x(t1 ) with x(t1 ) = x(t1 )}.
And the space P C[0, 1] is a Banach space with norm
kxkP C = sup{|x(t)| : t ∈ [0, 1]}.
Lemma 2.1. The linear impulsive boundary value problem
D α u(t) − λu(t) = σ(t),
t ∈ (0, 1] \ {t1 },
(2.1)
lim t1−α u(t) = u(1),
(2.2)
lim (t − t1 )1−α (u(t) − u(t1 )) = a1 ,
(2.3)
t→0+
t→t+
1
where λ, a1 ∈ R are constants and σ ∈ C[0, 1], has a unique solution u ∈ P C1−α [0, 1] given by
u(t) =
1
Z
Gλ,α (t, s)σ(s)ds + Γ(α)Gλ,α (t, t1 )a1 ,
0
(2.4)
where
Gλ,α (t, s)
=
Γ(α)Eα,α (λtα )Eα,α (λ(1−s)α )tα−1 (1−s)α−1
+
1−Γ(α)Eα,α (λ)
Γ(α)Eα,α (λtα )Eα,α (λ(1−s)α )tα−1 (1−s)α−1
,
1−Γ(α)Eα,α (λ)
λk
k=0 Γ((k+1)α)
P∞
and Eα,α (λ) =
(t − s)α−1 Eα,α (λ(t − s)α ), 0 ≤ s ≤ t ≤ 1,
0 ≤ t < s ≤ 1,
is Mittag-Leffler function (see [6]).
Proof. By Theorem 3.2 in [2], we have that
u
¯(t) =
Z
0
1
Gλ,α (t, s)σ(s)ds,
t ∈ (0, 1] \ {t1 }
(2.5)
EJQTDE, 2011 No. 3, p. 3
is a unique solution of the linear problem (2.1)-(2.2), and u¯ ∈ P C1−α [0, 1]. Set
w(t) = Gλ,α (t, t1 )Γ(α)a1 ,
t ∈ [0, 1].
For each t < t1 , we have
D α w(t) =
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α α−1
D (t
Eα,α (λtα ))
1 − Γ(α)Eα,α (λ)
∞
λi−1 αi−1
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α X
D
t
=
.
1 − Γ(α)Eα,α (λ)
Γ(αi)
i=1
!
Using the identities
D α tµ =
Γ(µ + 1) µ−α
t
(µ > −1),
Γ(µ + 1 − α)
Dα tα−1 = 0,
we get
D α w(t) =
∞
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 X
λi−1
tα(i−1)−1
1 − Γ(α)Eα,α (λ)
Γ(α(i
−
1))
i=2
=λ
∞
(−λ)i−1 αi−1
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 X
t
1 − Γ(α)Eα,α (λ)
Γ(αi)
i=1
=λ
∞
(λtα )i
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α−1 X
t
1 − Γ(α)Eα,α (λ)
Γ(αi + α)
i=0
=λ
Γ(α)Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 tα−1 Eα,α (λtα )
Γ(α)a1
1 − Γ(α)Eα,α (λ)
= λGα,α (t, t1 )Γ(α)a1 = λw(t),
t < t1 .
Similarly, we can obtain that
D α w(t) − λw(t) = 0,
t > t1 .
Thus, we have Dα w(t) − λw(t) = 0 for t ∈ (0, 1] \ {t1 }. Moreover, we have
lim t1−α w(t) − w(1)
t→0+
= lim+ t1−α
t→0
Γ(α)2 Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 α−1
t
Eα,α (λtα )
1 − Γ(α)Eα,α (λ)
!
Γ(α)Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 Eα,α (λ)
−
+ (1 − t1 )α−1 Eα,α (λ(1 − t1 )α ) Γ(α)a1
1 − Γ(α)Eα,α (λ)
=
Γ(α)Eα,α (λ(1 − t1 )α )(1 − t1 )α−1 a1 (1 − t1 )α−1 Eα,α (λ(1 − t1 )α )Γ(α)a1
−
1 − Γ(α)Eα,α (λ)
1 − Γ(α)Eα,α (λ)
EJQTDE, 2011 No. 3, p. 4
= 0,
and
lim+ (t − t1 )1−α (w(t) − w(t1 ))
t→t1
= lim (t − t1 )1−α · (t − t1 )α−1 Eα,α (λ(t − t1 )α )Γ(α)a1
t→t+
1
α
= lim Eα,α (λ(t − t1 ) )Γ(α)a1 = lim
t→t+
1 i=0
t→t+
1
=
1
Γ(α) Γ(α)a1
∞
X
(λ(t − t1 )α )i
Γ(αi + α)
Γ(α)a1
= a1 .
In consequence,
u(t) = u
¯(t) + w(t) =
Z
1
0
Gλ,α (t, s)σ(s)ds + Γ(α)Gλ,α (t, t1 )a1
is the solution of problem (2.1)-(2.3).
Next we prove that the solution of BVP (2.1)-(2.3) is unique. Suppose that u1 , u2 ∈
P C1−α [0, 1] are two solutions of BVP (2.1)-(2.3). Let v = u1 − u2 , then we have
D α v(t) − λv(t) = 0,
t ∈ (0, 1] \ {t1 },
(2.6)
lim t1−α v(t) − v(1) = 0,
(2.7)
lim (t − t1 )1−α (v(t) − v(t1 )) = 0.
(2.8)
t→0+
t→t+
1
By (2.5) and (2.6)-(2.7), we get that v(t) ≡ 0 for any t ∈ (0, 1] \ {t1 }. Since v ∈ P C1−α [0, 1],
v(t1 ). On the other hand, limt→t− t1−α v(t) = 0. Thus, we
we have limt→t− t1−α v(t) = t1−α
1
1
1
obtain that v(t1 ) = 0. Hence, u1 (t) = u2 (t) for each t ∈ (0, 1]. Moreover, by (2.7), we
have limt→0+ t1−α v(t) = v(1) = 0, which implies that limt→0+ t1−α u1 (t) = limt→0+ t1−α u2 (t).
Therefore, u1 = u2 .
Remark 2.2. For α = 1, Lemma 2.1 reduces to the one for a first order linear impulsive
boundary value problem
3. Main results
Lemma 3.1. Suppose that
(H1 ) there exist positive constants M and m such that
EJQTDE, 2011 No. 3, p. 5
|f (t, u)| ≤ M,
|I(u)| ≤ m,
∀t ∈ [0, 1], u ∈ R,
(3.1)
holds, then the operator A : P C1−α [0, 1] → P C1−α [0, 1] is well defined.
Proof. Define the operator A as follows :
(Ax)(t) :=
Z
1
0
Gλ,α (t, s)f (s, x(s))ds + Γ(α)Gλ,α (t, t1 )I(x(t1 )).
(3.2)
By Lemma 2.1, it is easy to see that a function x is a solution to (1.1)-(1.3) if and only
if x is a fixed point of A. Using (H1 ), we check that t1−α (Au)(t)|[0,t1 ] ∈ C[0, t1 ] and (t −
t1 )1−α (Au)(t)|(t1 ,1] ∈ C(t1 , 1]. Here, we only prove that (t − t1 )1−α (Au)(t)|(t1 ,1] ∈ C(t1 , 1].
Similarly, we can prove that t1−α (Au)(t)|[0,t1 ] ∈ C[0, t1 ]. For every u ∈ P C1−α [0, 1], for any
τ1 , τ2 ∈ (t1 , 1] with τ1 < τ2 , we have
(τ1 − t1 )1−α (Au)(τ1 ) − (τ2 − t1 )1−α (Au)(τ2 )
Z
≤ (τ1 − t1 )1−α
1
0
1−α
Gλ,α (τ1 , s)f (s, u(s))ds − (τ2 − t1 )
Z
0
1
Gλ,α (τ2 , s)f (s, u(s))ds
+ (τ1 − t1 )1−α Γ(α)Gλ,α (τ1 , t1 )I(u(t1 )) − (τ2 − t1 )1−α Γ(α)Gλ,α (τ2 , t1 )I(u(t1 ))
Z
1 Γ(α)E (λ(1 − s)α )(1 − s)α−1
α,α
=
0
1 − Γ(α)Eα,α (λ)
×[(τ1 − t1 )1−α Eα,α (λτ1α )τ1α−1 − (τ2 − t1 )1−α Eα,α (λτ2α )τ2α−1 ]f (s, u(s))ds
+
Z
τ1
0
[(τ1 − t1 )1−α (τ1 − s)α−1 Eα,α (λ(τ1 − s)α )
−(τ2 − t1 )1−α (τ2 − s)α−1 Eα,α (λ(τ2 − s)α )]f (s, u(s))ds
−
Z
τ2
τ1
≤
(τ2 − t1 )1−α (τ2 − s)α−1 Eα,α (λ(τ2 − s)α f (s, u(s))ds
+ [(τ1 − t1 )1−α Gλ,α (τ1 , t1 ) − (τ2 − t1 )1−α Gλ,α (τ2 , t1 )]Γ(α)I(u(t1 ))
Γ(α)M
|(τ1 − t1 )1−α Eα,α (λτ1α )τ1α−1 − (τ2 − t1 )1−α Eα,α (λτ2α )τ2α−1 |
|1 − Γ(α)Eα,α (λ)|
×
Z
0
1
Eα,α (λ(1 − s)α )(1 − s)α−1 ds
+M ((τ2 − t1 )1−α − (τ1 − t1 )1−α )
+M (τ2 − t1 )1−α
Z
0
τ1
Z
τ1
0
(τ1 − s)α−1 Eα,α (λ(τ1 − s)α )ds
|(τ1 − s)α−1 Eα,α (λ(τ1 − s)α ) − (τ2 − s)α−1 Eα,α (λ(τ2 − s)α )|ds
EJQTDE, 2011 No. 3, p. 6
+M (τ2 − t1 )1−α
+
τ2
Z
τ1
(τ2 − s)α−1 Eα,α (λ(τ2 − s)α )ds
Γ2 (α)mEα,α (λ(1 − t1 )α )(1 − t1 )1−α
|1 − Γ(α)Eα,α (λ)|
×|(τ1 − t1 )1−α τ1α−1 Eα,α (λτ1α ) − (τ2 − t1 )1−α τ2α−1 Eα,α (λτ2α )|
+|Eα,α (λ(τ1 − t1 )α ) − Eα,α (λ(τ2 − t1 )α )|
→ 0,
as τ1 → τ2 ,
since
Z
1
Z
τ1
Z
τ2
Z
τ1
0
Eα,α (λ(1 − s)α )(1 − s)α−1 ds = Eα,α+1 (λ) ≤ Eα,α+1 (|λ|),
0
τ1
and
0
≤
Eα,α (λ(τ1 − s)α )(τ1 − s)α−1 ds = τ1α Eα,α+1 (λτ1α ) ≤ τ1α Eα,α+1 (|λ|τ1α ),
Eα,α (λ(τ2 − s)α )(τ2 − s)α−1 ds = (τ2 − τ1 )α Eα,α+1 (λ(τ2 − τ1 )α ),
|(τ1 − s)α−1 Eα,α (λ(τ1 − s)α ) − (τ2 − s)α−1 Eα,α (λ(τ2 − s)α )|ds
∞
X
i=0
|λ|i
Γ(αi + α)
Z
τ1
0
(τ1 − s)αi+α−1 − (τ2 − s)αi+α−1 ds
tend to zero as τ1 → τ2 by (4.13) and (4.14) in [2].
So, (t − t1 )1−α (Au)(t)|(t1 ,1] ∈ C(t1 , 1]. On the other hand, we have limt→t− (Au)(t) and
1
lim (t − t1 )1−α (Au)(t)
t→t+
1
= lim (t − t1 )1−α
t→t+
1
+
Z
0
t
"Z
1
0
Γ(α)Eα,α (λtα )Eα,α (λ(1 − s)α )tα−1 (1 − s)α−1
f (s, u(s))ds
1 − Γ(α)Eα,α (λ)
(t − s)α−1 Eα,α (λ(t − s)α )f (s, u(s))ds
+Γ(α)
Γ(α)Eα,α (λtα )Eα,α (λ(1 − t1 )α )tα (1 − t1 )α−1
1 − Γ(α)Eα,α (λ)
i
+(t − t1 )α−1 Eα,α (λ(t − t1 )α ) I(u(t1 ))
= lim (t − t1 )1−α · Γ(α)(t − t1 )α−1 Eα,α (λ(t − t1 )α )I(u(t1 ))
t→t+
1
EJQTDE, 2011 No. 3, p. 7
= I(u(t1 ))
exist. Thus, A : P C1−α [0, 1] → P C1−α [0, 1].
For convenience, set
M1 :=
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1 tα1
(Γ(α))2
+ tα1 Eα,α (|λ|tα1 )
,
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
(3.3)
M2 :=
(Γ(α))3 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )(1 − t1 )2α−1
,
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.4)
M3 :=
M4 :=
Γ(α)(Eα,α (|λ|))2 t12α−1
|1−Γ(α)(Eα,α (λ)|α
Γ(α)(Eα,α (|λ|))2 t12α−1
|1−Γ(α)Eα,α (λ)|α
2
1
+ t2α−1
(1 − t1 )1−α Eα,α (|λ|) (Γ(α))
1
Γ(2α) , if 0 < α ≤ 2 ,
2
+ (1 − t1 )1−α Eα,α (|λ|) (Γ(α))
Γ(2α) ,
if
1
2
< α ≤ 1,
(Γ(α))3 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )tα−1
(1 − t1 )α
1
,
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.5)
(3.6)
N1 :=
(1 − t1 )α−1
(Γ(α))2 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )tα−1
1
,
|1 − Γ(α)Eα,α (λ)|
(3.7)
N2 :=
(Γ(α))2 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )t2α−2
1
+ Γ(α)Eα,α (|λ|(1 − t1 )α )tα−1
.
1
|1 − Γ(α)Eα,α (λ)|
(3.8)
Theorem 3.2. Suppose that (H1 ) and the following condition hold:
(H2 ) there exist positive constant k, and l such that
|f (t, u) − f (t, v)| ≤ k|u − v|,
|I(u) − I(v)| ≤ l|u − v|,
∀t ∈ [0, 1], u, v ∈ R.
(3.9)
Then the problem (1.1)-(1.3) has a unique solution in P C1−α [0, 1], provided that
max{k(M1 + M2 ) + lN1 , k(M3 + M4 ) + lN2 } < 1.
(3.10)
Proof. By (H1 ) and Lemma 3.1, we have A : P C1−α [0, 1] → P C1−α [0, 1]. For any x, y ∈
P C1−α [0, 1], and each t ∈ [0, t1 ], we obtain by (H2 ) that
t
1−α
|(Ax)(t) − (Ay)(t)| ≤
Z
1
0
t1−α |Gλ,α (t, s)||f (s, x(s)) − f (s, y(s))|ds
+t1−α Γ(α)|Gλ,α (t, t1 )||I(x(t1 )) − I(y(t1 ))|
≤
Z
1
0
=k
Z
t1−α |Gλ,α (t, s)|k|x(s) − y(s)|ds + t1−α Γ(α)|Gλ,α (t, t1 )|l|x(t1 ) − y(t1 )|
0
t1
t1−α |Gλ,α (t, s)|sα−1 s1−α |x(s) − y(s)|ds
EJQTDE, 2011 No. 3, p. 8
1
Z
+k
t1−α |Gλ,α (t, s)|(s − t1 )α−1 (s − t1 )1−α |x(s) − y(s)|ds
t1
+lΓ(α)t1−α |Gλ,α (t, t1 )|tα−1
t11−α |x(t1 ) − y(t1 )|
1
Z
≤ kx − yk1−α k
+k
1
Z
t
t1
1−α
t1
0
t1−α |Gλ,α (t, s)|sα−1 ds
α−1
|Gλ,α (t, s)|(s − t1 )
ds + lΓ(α)t
1−α
|Gλ,α (t, t1 )|tα−1
1
(3.11)
.
From the expression of Gλ,α (t, s), and (4.21) in [2], we have
Z
t1
t
1−α
0
|Gλ,α (t, s)|s
+
Z
t
0
≤
α−1
ds ≤
Z
t1
t
0
1−α Γ(α)Eα,α (λt
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1 α−1
ds
s
1 − Γ(α)Eα,α (λ)
t1−α (t − s)α−1 |Eα,α (λ(t − s)α )|sα−1 ds
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1
|1 − Γ(α)Eα,α (λ)|
+t1−α Eα,α (|λ|tα1 )
Z
t
Z
t1
sα−1 ds
0
(t − s)α−1 sα−1 ds
0
=
(Γ(α))2
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1 tα1
+ tα Eα,α (|λ|tα1 )
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
≤
(Γ(α))2
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|)(1 − t1 )α−1 tα1
+ tα1 Eα,α (|λ|tα1 )
= M1 .
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
(3.12)
And
Z
1
t1
t1−α |Gλ,α (t, s)|(s − t1 )α−1 ds
≤
Z
1
t
t1
≤
=
1−α Γ(α)Eα,α (λt
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1
(s − t1 )α−1 ds
1 − Γ(α)Eα,α (λ)
Γ(α)Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )
|1 − Γ(α)Eα,α (λ)|
Z
1
t1
(1 − s)α−1 (s − t1 )α−1 ds
(Γ(α))3 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )(1 − t1 )2α−1
= M2 ,
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.13)
since
Z
1
t1
α−1
(1 − s)
α−1
(s − t1 )
ds =
Z
0
1−t1
sα−1 (1 − t1 − s)α−1 ds = (1 − t1 )2α−1
(Γ(α))2
.
Γ(2α)
(3.14)
EJQTDE, 2011 No. 3, p. 9
Moreover, we have
Γ(α)t1−α |Gλ,α (t, t1 )|tα−1
1
≤
(1 − t1 )α−1
(Γ(α))2 Eα,α (|λ|tα1 )Eα,α (|λ|(1 − t1 )α )tα−1
1
= N1 .
|1 − Γ(α)Eα,α (λ)|
(3.15)
Substituting (3.12), (3.13) and (3.15) into (3.11), we get
sup t1−α |(Ax)(t) − (Ay)(t)| ≤ (k(M1 + M2 ) + N1 )kx − yk1−α .
(3.16)
t∈[0,t1 ]
On the other hand, for each x, y ∈ P C1−α [0, 1] and t ∈ (t1 , 1], we obtain that
(t − t1 )1−α |(Ax)(t) − (Ay)(t)|
≤
1
Z
0
(t − t1 )1−α |Gλ,α (t, s)||f (s, x(s)) − f (s, y(s))|ds
+(t − t1 )1−α Γ(α)|Gλ,α (t, t1 )||I(x(t1 )) − I(y(t1 ))|
≤
1
Z
0
(t − t1 )1−α |Gλ,α (t, s)|k|x(s) − y(s)|ds
+(t − t1 )1−α Γ(α)|Gλ,α (t, t1 )|l|x(t1 ) − y(t1 )|
=k
Z
t1
0
+k
(t − t1 )1−α |Gλ,α (t, s)|sα−1 s1−α |x(s) − y(s)|ds
1
Z
t1
(t − t1 )1−α |Gλ,α (t, s)|(s − t1 )α−1 (s − t1 )1−α |x(s) − y(s)|ds
+(t − t1 )1−α Γ(α)l|Gλ,α (t, t1 )|tα−1
t1−α
|x(t1 ) − y(t1 )|
1
1
Z
≤ kx − yk1−α k
+k
Z
1
t1
0
t1
(t − t1 )1−α |Gλ,α (t, s)|sα−1 ds
(t − t1 )1−α |Gλ,α (t, s)|(s − t1 )α−1 ds + lΓ(α)|Gλ,α (t, t1 )|(t − t1 )1−α tα−1
.
1
(3.17)
By (4.21) in [2], we have
Z
0
t1
(t − t1 )1−α |Gλ,α (t, s)|sα−1 ds
≤
Z
t1
Z
t
0
+
0
1−α Γ(α)Eα,α (λt
(t − t1 )
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1 α−1
ds
s
1 − Γ(α)Eα,α (λ)
(t − t1 )1−α (t − s)α−1 |Eα,α (λ(t − s)α )|sα−1 ds
EJQTDE, 2011 No. 3, p. 10
≤ (1 − t1 )1−α
Γ(α)(Eα,α (|λ|))2 tα−1
(1 − t1 )α−1
1
|1 − Γ(α)Eα,α (λ)|
+(1 − t1 )1−α Eα,α (|λ|)
Z
t
Z
t1
sα−1 ds
0
(t − s)α−1 sα−1 ds
0
=
(Γ(α))2
Γ(α)(Eα,α (|λ|))2 t2α−1
1
+ (1 − t1 )1−α Eα,α (|λ|)t2α−1
|1 − Γ(α)Eα,α (λ)|α
Γ(2α)
≤
Γ(α)(Eα,α (|λ|))2 t2α−1
1
|1 − Γ(α)Eα,α (λ)|α
t2α−1 (1 − t1 )1−α Eα,α (|λ|) (Γ(α))2 , if 0 < α ≤ 1 ,
1
2
Γ(2α)
+
2
1
(1 − t1 )1−α Eα,α (|λ|) (Γ(α)) ,
if 2 < α ≤ 1,
Γ(2α)
= M3 .
(3.18)
From (3.14), we get
Z
1
t1
(t − t1 )1−α |Gλ,α (t, s)|(s − t1 )α−1 ds
=
Z
1
t1
(t − t1 )
≤ (1 − t1 )1−α
=
1−α Γ(α)Eα,α (λt
α )E
α,α (λ(1
− s)α )tα−1 (1 − s)α−1
(s − t1 )α−1 ds
1 − Γ(α)Eα,α (λ)
Γ(α)Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )tα−1
1
|1 − Γ(α)Eα,α (λ)|
Z
1
t1
(s − t1 )α−1 (1 − s)α−1 ds
(Γ(α))3 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )tα−1
(1 − t1 )α
1
= M4 .
|1 − Γ(α)Eα,α (λ)|Γ(2α)
(3.19)
Moreover, we have
Γ(α)|Gλ,α (t, t1 )|(t − t1 )1−α tα−1
1
≤
(Γ(α))2 Eα,α (|λ|tα )Eα,α (|λ|(1 − t1 )α )tα−1 (1 − t1 )α−1 (t − t1 )1−α tα−1
1
|1 − Γ(α)Eα,α (λ)|
+Γ(α)tα−1
Eα,α (|λ|(t − t1 )α )
1
≤
(Γ(α))2 Eα,α (|λ|)Eα,α (|λ|(1 − t1 )α )t2α−2
1
|1 − Γ(α)Eα,α (λ)|
+Γ(α)Eα,α (|λ|(1 − t1 )α )tα−1
= N2 .
1
(3.20)
Substituting (3.18)-(3.20) into (3.17), we obtain
sup (t − t1 )1−α |(Ax)(t) − (Ay)(t)| ≤ (k(M3 + M4 ) + lN )kx − yk1−α .
(3.21)
t∈(t1 ,1]
EJQTDE, 2011 No. 3, p. 11
From (3.11), (3.16) and (3.21), we have
kAx − Ayk1−α
= max
(
sup t1−α |(Ax)(t) − (Ay)(t)|,
t∈[0,t1 ]
)
sup (t − t1 )1−α |(Ax)(t) − (Ay)(t)|
t∈(t1 ,1]
≤ max{k(M1 + M2 ) + lN1 , k(M3 + M4 ) + lN2 }kx − yk1−α ,
(3.22)
which implies that A is a contraction (by (3.10)). Therefore it has a unique fixed point.
4. An example
The following illustrative example will demonstrate the effectiveness of our main result.
Example 4.1. Consider the nonlinear impulsive fractional differential equation with periodic
boundary conditions as follows:
1
t2
D 0.9 u(t) − u(t) = 2
arctan u(t),
3
t + 11
t ∈ (0, 1] \ {0.5},
(4.1)
lim t0.1 u(t) = u(1),
(4.2)
t→0+
lim (t − 0.5)0.1 (u(t) − u(0.5)) =
t→0.5+
1
sin(u(0.5)).
13
(4.3)
Then BVP (4.1)-(4.3) reduces to BVP (1.1)-(1.3) with α = 0.9, λ = 13 , f (t, u) =
I(u) =
1
13
t2
t2 +11
arctan u,
sin u and t1 = 0.5.
Obviously, there exist positive constants M =
|f (t, u)| ≤ M,
|I(u)| ≤ m,
π
24 ,
m=
1
13 ,
k=
1
12
and l =
1
13
such that
t ∈ [0, 1], u ∈ R,
and
|f (t, u) − f (t, v)| ≤ k|u − v|,
|I(u) − I(v)| ≤ l|u − v|,
t ∈ [0, 1], u, v ∈ R.
So that the conditions (H1 ) and (H2 ) hold. Moreover, we have
Eα,α (λ) = E0.9,0.9
1
3
=
1
1
1
1
1
+
+ 2
+ 3
+ 4
+ ···
Γ(0.9) 3Γ(1.8) 3 Γ(2.7) 3 Γ(3.6) 3 Γ(4.5)
Thus, we obtain
E0.9,0.9
1
3
>
1
1
1
+
+ 2
= 1.3656,
Γ(0.9) 3Γ(1.8) 3 Γ(2.7)
(4.4)
and
EJQTDE, 2011 No. 3, p. 12
E0.9,0.9
1
3
1
1
1
1
+
+ 2
+
<
Γ(0.9) 3Γ(1.8) 3 Γ(2.7) 3
1
1
+ 4 + ···
3
3
3
= 1.3656 + 0.0185 = 1.3841.
(4.5)
Similarly, we have
Eα,α (|λ|tα1 )
α
= Eα,α (|λ|(1 − t1 ) ) = E0.9,0.9
1
3 · 20.9
< 1.149.
(4.6)
By (4.4)-(4.6) and simple calculation, we can obtain the estimation of M1 , M2 , M3 , M4 , N1 and
N2 (are as in (3.3)-(3.8) as follows :
M1 < 3.019,
M2 < 1.9351,
M4 < 2.4333,
N1 < 3.6153,
M3 < 4.2091,
N2 < 5.6709.
Hence, we get
max{k(M1 + M2 ) + lN1 , k(M3 + M4 ) + lN2 } = 0.9898 < 1,
that is (3.10) holds. So, it follows from Theorem 3.2 that the BVP (4.1)-(4.3) has a unique
solution in P C0.1 [0, 1].
Acknowledgements
The authors would like to thank the referee for their valuable suggestions and comments.
This work is supported by the National Natural Science Foundation of China (10771212).
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